Discuss existence and uniqueness for a Cauchy problem I don't know what's happening with this exercise. I need a help becuase I'm quite puzzled.

Consider the Cauchy problem
\begin{cases}
y'=\frac{2}{t} y + 2 t \sqrt{y} \\
y(1)=0
\end{cases}
Study the existence and uniqueness

Here $$f(t,y)=\frac{2}{t} y + 2 t \sqrt{y}$$ Since $y\geq0$ (I have the square root), I consider as open neigbourhood $K = \{t: |t-1|< r_1 \} \times \{y: 0 < y < r_2 \}$, but in this way I am in trouble with $$f_y(t,y)= \frac{2}{t} + \frac{t}{\sqrt{y}}$$ because it's discontinuous at $y=0$.
So I should look for a weaker condition as Lipschitz continuity: I take $(t,y_1)$ and $(t,y_2)$ in $K$:
$$|\frac{2}{t} \bigl(y_1 - y_2 \bigr) + 2t \bigl( \sqrt{y_1} - \sqrt{y_2} \bigr)|  \leq |\frac{2}{t} \bigl(y_1 - y_2 \bigr)| + |2t \bigl( \sqrt{y_1} - \sqrt{y_2} \bigr)| $$
but the second term of the inequality is quite problematic: it is like proving that $x \mapsto \sqrt{x}$ is Lipschitz for $x\geq0$, which is known to be false.

So, I can't apply the theorem actually...Am I wrong? If so, what are my mistakes?
 A: The r.h.s. $f(t,y)$, defined in $\Omega := (0, +\infty) \times [0,+\infty)$, is continuous in $\Omega$ but it is not locally Lipschitz continuous.
Hence, Peano's theorem guarantees local existence, but uniqueness need not hold (and, indeed, in our case we have more than one solution).
Further, $f$ is sublinear in $y$, meaning that $|f(t,y)| \leq a(t) + b(t) |y|$ for some continuous functions $a, b \in C((0,+\infty))$, so that all solutions are global (meaning that every solution admits an extension on $(0,+\infty)$).
Let us compute the solutions of the given Cauchy problem.
One solution is the constant function $y(t) = 0$, $t\in (0,+\infty)$.
Other solutions bifurcate from the constant solution at some time $\tau \geq 1$.
In order to find them, let us first compute the strictly positive solutions of the differential equation.
With the change of variable $z = \sqrt{y}$ we are let to the linear equation
$z' = z/t + t$, whose solutions are of the form $z(t) = ct + t^2$, for some constant $c\in \mathbb{R}$.
Recall that we are interested only in positive solutions defined in some subinterval of $(0,+\infty)$. The corresponding $y$ are then of the form
$$
y_\tau (t) = t^2 (t- \tau)^2, \qquad t > \max\{\tau, 0\},
$$
where $\tau$ is a real parameter.
It is easily seen that, if $\tau \geq 1$, then $y_\tau$ can be prolonged to the left with the $0$ solution, obtaining the global solution of the Cauchy problem
$$
y_\tau(t) :=
\begin{cases}
0, & \text{if}\ t \in (0, \tau],
\\
t^2(t-\tau)^2, & \text{if}\ t > \tau\,.
\end{cases}
$$
In conclusion, for every $\tau \geq 1$ the above function is a solution of the Cauchy problem. (This family of solutions is called the Peano brush.)
